Question

Use the result of Problem 54 to show that in any elastic collision between two objects, the relative speed of the two is the same before and after the collision. [Hints: Look at the collision in its \(\mathrm{CM}\) frame - the reference frame in which the \(\mathrm{CM}\) is at rest. The relative speed of two objects is the same in any inertial reference frame.]

Short answer

Answer: Yes, in an elastic collision between two objects, the relative speed of the objects remains the same before and after the collision.

Step-by-step solution

Define the variables and given information

Let's denote the two objects as A and B, with masses \(m_A\) and \(m_B\) respectively. Their velocities before the collision in the CM frame are \(v_{A, i}\) and \(v_{B, i}\), and after the collision, they are \(v_{A, f}\) and \(v_{B, f}\). We also define the initial and final relative speeds as \(v_{rel, i} = |v_{A, i} - v_{B, i}|\) and \(v_{rel, f} = |v_{A, f} - v_{B, f}|\).

Apply conservation of momentum in the CM frame

Since the total momentum is conserved, we can write the following equation for the CM frame: \(m_A v_{A, i} + m_B v_{B, i} = m_A v_{A, f} + m_B v_{B, f}\) Also, in the CM frame, the initial total momentum is zero: \(m_A v_{A, i} + m_B v_{B, i} = 0\)

Apply conservation of kinetic energy in the CM frame

In an elastic collision, the kinetic energy of the system is also conserved. Therefore, we can write the following equation in the CM frame: \(\frac{1}{2} m_A v_{A, i}^2 + \frac{1}{2} m_B v_{B, i}^2 = \frac{1}{2} m_A v_{A, f}^2 + \frac{1}{2} m_B v_{B, f}^2\)

Combine the conservation equations

Square the conservation of momentum equation and substitute it into the conservation of kinetic energy equation: \(m_A^2 v_{A, i}^2 + 2m_A m_B v_{A, i} v_{B, i} + m_B^2 v_{B, i}^2 = m_A^2 v_{A, f}^2 + 2m_A m_B v_{A, f} v_{B, f} + m_B^2 v_{B, f}^2\) Since the left side and right side of the equation represent the initial and final kinetic energy respectively, we can rewrite it as: \(2m_A m_B v_{A, i} v_{B, i} + m_A^2 v_{A, i}^2 + m_B^2 v_{B, i}^2 = 2m_A m_B v_{A, f} v_{B, f} + m_A^2 v_{A, f}^2 + m_B^2 v_{B, f}^2\) Now, we can cancel out the terms with \(m_A^2\) and \(m_B^2\) on both sides, which yields: \(2m_A m_B (v_{A, i} v_{B, i} - v_{A, f} v_{B, f}) = 0\)

Find the relation between initial and final relative speeds

Since the masses of the objects are not zero, we can divide both sides of the equation by \(2m_A m_B\) and obtain: \(v_{A, i} v_{B, i} = v_{A, f} v_{B, f}\) We will now write the initial and final relative speeds in terms of the velocities of A and B: \(v_{rel, i}^2 = (v_{A, i} - v_{B, i})^2\) \(v_{rel, f}^2 = (v_{A, f} - v_{B, f})^2\) Subtract these two equations: \(v_{rel, i}^2 - v_{rel, f}^2 = (v_{A, i}^2 - 2v_{A, i} v_{B, i} + v_{B, i}^2) - (v_{A, f}^2 - 2v_{A, f} v_{B, f} + v_{B, f}^2)\) From the conservation of momentum equation, we know that \(v_{A, i} v_{B, i} = v_{A, f} v_{B, f}\). Therefore: \(v_{rel, i}^2 - v_{rel, f}^2 = (v_{A, i}^2 + v_{B, i}^2) - (v_{A, f}^2 + v_{B, f}^2)\)

Conclusion

From the conservation of kinetic energy equation, we know that the initial and final total kinetic energy are equal. So, the difference between the initial and final relative speeds must be zero: \(v_{rel, i}^2 - v_{rel, f}^2 = 0\) This implies that: \(v_{rel, i} = v_{rel, f}\) Thus, in any elastic collision between two objects, the relative speed of the two is the same before and after the collision. This result is valid in any inertial reference frame.